OPERATIONS MANAGEMENT: Creating Value Along the Supply Chain, Canadian Edition Robert S. Russell, Bernard W. Taylor III, Ignacio Castillo, Navneet Vidyarthi CHAPTER 14 SUPPLEMENT Linear Programming

Lecture Outline Model Formulation Graphical Solution Method Linear Programming Model Solution Solving Linear Programming Problems with Excel Sensitivity Analysis

Linear Programming (LP) A model consisting of linear relationships representing a firm’s objective and resource constraints A mathematical modeling technique which determines a level of operational activity in order to achieve an objective, subject to restrictions called constraints

Types of LP

Types of LP

Types of LP

LP Model Formulation Decision variables Objective function Constraint symbols representing levels of activity of an operation Objective function linear relationship for the objective of an operation most frequent business objective is to maximize profit most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost Constraint linear relationship representing a restriction on decision making

LP Model Formulation Max/min z = c1x1 + c2x2 + ... + cnxn subject to: a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1 a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2 : an1x1 + an2x2 + ... + annxn (≤, =, ≥) bn xj = decision variables bi = constraint levels cj = objective function coefficients aij = constraint coefficients Constraints

Highlands Craft Store Labor Clay Revenue Product (hr/unit) (lb/unit) ($/unit) Bowl 1 4 40 Mug 2 3 50 There are 40 hours of labor and 120 pounds of clay available each day Decision variables x1 = number of bowls to produce x2 = number of mugs to produce Resource Requirements

Highlands Craft Store Maximize Z = $40 x1 + 50 x2 Subject to x1 + 2x2 40 hr (labor constraint) 4x1 + 3x2 120 lb (clay constraint) x1 , x2 0 Solution is x1 = 24 bowls x2 = 8 mugs Revenue = $1,360

Graphical Solution Method Plot model constraint on a set of coordinates in a plane Identify the feasible solution space on the graph where all constraints are satisfied simultaneously Plot objective function to find the point on boundary of this space that maximizes (or minimizes) value of objective function

Graphical Solution Method 4 x1 + 3 x2 120 lb x1 + 2 x2 40 hr Area common to both constraints 50 – 40 – 30 – 20 – 10 – 0 – | 10 60 50 20 30 40 x1 x2 Objective function

Computing Optimal Values 40 – 30 – 20 – 10 – 0 – x1 + 2x2 = 40 4x1 + 3x2 = 120 4x1 + 8x2 = 160 -4x1 - 3x2 = -120 5x2 = 40 x2 = 8 x1 + 2(8) = 40 x1 = 24 4 x1 + 3 x2 =120 lb x1 + 2 x2 =40 hr | 10 20 30 40 x1 x2 Z = $40(24) + $50(8) = $1,360 24 8

Extreme Corner Points x1 = 0 bowls x2 =20 mugs Z = $1,000 x2 A B C | 20 30 40 10 x1 x2 40 – 30 – 20 – 10 – 0 –

Objective Function 4x1 + 3x2 =120 lb Z = 70x1 + 20x2 Optimal point: 40 – 30 – 20 – 10 – 0 – 4x1 + 3x2 =120 lb x1 + 2x2 =40 hr B | 10 20 30 40 x1 x2 C A Z = 70x1 + 20x2 Optimal point: x1 = 30 bowls x2 =0 mugs Z = $2,100

Minimization Problem CHEMICAL CONTRIBUTION Brand Nitrogen (lb/bag) Phosphate (lb/bag) Gro-plus 2 4 Crop-fast 4 3 Minimize Z = $6x1 + $3x2 subject to 2x1 + 4x2  16 lb of nitrogen 4x1 + 3x2  24 lb of phosphate x1, x2  0

Graphical Solution x1 = 0 bags of Gro-plus x2 = 8 bags of Crop-fast 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – | 2 4 6 8 10 12 14 x1 x2 A B C x1 = 0 bags of Gro-plus x2 = 8 bags of Crop-fast Z = $24 Z = 6x1 + 3x2

Simplex Method Mathematical procedure for solving LP problems Follow a set of steps to reach optimal solution Slack variables added to ≤ constraints to represent unused resources x1 + 2x2 + s1 = 40 hours of labor 4x1 + 3x2 + s2 = 120 lb of clay Surplus variables subtracted from ≥ constraints to represent excess above resource requirement. 2x1 + 4x2 ≥ 16 is transformed into 2x1 + 4x2 - s1 = 16 Slack/surplus variables have a 0 coefficient in the objective function Z = $40x1 + $50x2 + 0s1 + 0s2

Solution Points With Slack Variables

Solution Points With Surplus Variables

Solving LP Problems with Excel Objective function =C6*B10+D6*B11 =E6-F6 =E7-F7 =C7*B10+D7*B11 Decision variables bowls (X1) = B10 mugs (x2) = B11 Click on “Data” to invoke “Solver”

Solving LP Problems with Excel After all parameters and constraints have been input, click on “Solve” Objective function Decision variables C6*B10+D6*B11≤40 and C7*B10+D7*B11≤120 Click on “Add” to insert constraints Click on “Options” to add non-negativity and linear conditions

LP Solution

Sensitivity Analysis Sensitivity range for labor; 30 to 80 lbs. Sensitivity range for clay; 60 to 160lbs. Shadow prices – marginal values – for labor and clay.

Sensitivity Range for Labor Hours

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